Geodesic equations in Schwarzschild geometry

Question

I am a little bit confused about the following equations. In the lecture we derived the following four equations for a geodesic motion of a particle in Schwarzschild geometry using the Lagrangian approach:

$(1)\hspace{10mm}\Theta = \frac{\pi}{2}$

$(2)\hspace{10mm}\big( 1- \frac{2\mu}{r}\big )\frac{\mathrm{d}t}{\mathrm{d}\tau}=K$

$(3)\hspace{10mm}r^{2}\frac{\mathrm{d}\varphi}{\mathrm{d}\tau} = h$

$(4)\hspace{10mm}c^{2}\big( 1- \frac{2\mu}{r}\big )\big(\frac{\mathrm{d}t}{\mathrm{d}\tau}\big)^{2}-\big( 1- \frac{2\mu}{r}\big )^{-1}\big(\frac{\mathrm{d}r}{\mathrm{d}\tau}\big)^{2}-r^{2}\big (\frac{\mathrm{d}\varphi}{\mathrm{d}\tau}\big )^2=\begin{cases}=c^{2}, \hspace{5mm}\text{massive particle}\\ = 0, \hspace{5mm}\text{massless particle}\end{cases}$

where $t,r,\theta,\varphi$ are the usual Schwarzschild coordinates and $K$ and $h$ are constant (they do not depend on any coordinate).

Now to my question: Because $K$ is a constant, which does not depend on any other coordinate, we can look at the limit $r\to\infty$. Using equation (2) at this limit we find trivially $K=1$ and because K doesnt depend on the radial coordinate, K has to be equal 1 everywhere. Therefore we can rewrite eq. (2) as

$$\frac{\mathrm{d}t}{\mathrm{d}\tau}=\bigg( 1- \frac{2\mu}{r}\bigg )^{-1}$$

But this cant be true!

Where is my thinking error?

Answer 1

First off, when you take the $r\to \infty$ limit in equation (2), you're assuming that the particle can actually reach infinity. Of course this is not always true, since there exist bound orbits in the Schwarzschild geometry. But assuming this is the case, it's also not true that $dt/d\tau = 1$ at infinity. From special relativity, $dt/d\tau$ is the time component of the four-velocity (also known as the energy), which is

$$u^t = \frac{1}{\sqrt{1-v^2}} = \gamma(v),$$

so the correct equation is

$$K = \gamma(v_\infty) \geq 1,$$

where $v_\infty$ is the speed at infinity. The possible values of $K$ then split into three regions: $K>1$ for the orbits that reach infinity (the analogs of hyperbolic orbits in the Kepler problem), $K<1$ for those that don't, and $K=1$ for the critical case of a trajectory whose velocity asymptotically approaches zero.

Answer 2

Your two constants $K$ and $h$ are not constant for all geodesics. They are constant for a particular geodesic. $K$ and $h$ are the energy and angular momentum of the geodesic. Each geodesic has fixed energy and angular momentum, but not all geodesics have the same energy and angular momentum. If you take the limit as $r\rightarrow\infty$, that result only applies to those geodesics that actually reach infinity, e.g. unbound geodesics.

@Javier also points out that if you want to consider $K(r\rightarrow\infty)$, you need to take the limit of the whole expression, including $dt/d\tau$.